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What is the sum expression for the sum of n and 5?
The sum of n and 5 is n+ 5
C code that will give the sum of an entered input?
#include<stdio.h> #include<conio.h> main() { int n,sum=0; printf("enter n value"); scanf("%d",n); while(n!=0) { n=n/10; sum=sum+n; n=n%10; } printf("sum is=%d",sum }
C program to find sum of 'n' number using function?
void main () { int no1, sum, n; clrscr() sum = 0; n = 1; printf("\n enter the number to which sum is to be generated"); scanf("%d",&no1); while(n<=no1) { sum=sum+n; n=n+1 } printf("\n the sum of %d = %d, no1, sum"); getch (); }
Recursive functions for the sum of squares?
int sum (int n){if (n
What is the c plus plus program for printing sum to n natural numbers?
main() { int i, n, sum=0; cout<<"Enter the limit"; cin>>n; for(i=0;i<=n;i++) sum=sum+i; cout<<"sum of "<<n<<" natural numbers ="<<sum; getch(); }
What represent the sum of n and the sum of 8 and 6?
The sum of n and the sum of 8 and 6 can be represented as n + (8 + 6). Simplifying the expression within the parentheses first, we get n + 14. Therefore, the final expression representing the sum of n and the sum of 8 and 6 is n + 14.
C program to compute the sum of digits of a given 5 digit number using for loop?
int SumOfDigits (int N) { int sum; if (N < 0) N = -N; for (sum = 0; N != 0; N /= 10) sum += N % 10; return sum; }
How write algorithm for finding the sum first ten even number?
Sum = 0 For N = 1 to 10 Sum = Sum + 2*N Next N Print Sum
Algorithm to find the sum of even numbers between 1 to 10?
sum = 0 for(n = 0; n <= 10; n += 2) sum += n;
Program to find sum of n numbers using recursion function?
int sum(n) { if (n==0) return 0; else return n+sum(n-1); }
What is the sum of n, n-1, n-2, and n-3?
The sum of n, n-1, n-2, and n-3 is 4n-6.
What is the sum of the whole numbers from 5 up to 100?
its 5035the summarian notation tells you that(sum of all #from 0 to a number'N'(sum of all #from 0 to N) = (n)+(n-1)+(n-2)+(n-3)+...+(2) +(1) or(sum of all #from 0 to N) = (1)+ (2) + (3) + (4)+...+(n-1)+(n)the two different sums are aligned by columns. now add the two colunms accordingly and you'll get2x(sum of all #from 0 to N)=(n+1)+(n+1)+...+(n+1) (n+1)is added n timesso2x(sum of all #from 0 to N) =n(n+1)(sum of all #from 0 to N) =n(n+1)/2so (sum of 5 to 100) = (sum of 0 to 100)- (sum of 0 to 5)=100(101)/2 - 5(6)/2 = 5050 - 15 = 5035
